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Calculus I · Functions and graphs

Learn to read function notation, identify the allowed inputs and possible outputs, and connect a formula to its graph.

Inputs, outputs and domain

A function assigns exactly one output to each allowed input. Its domain is the set of allowed inputs, and its range is the set of outputs it actually produces. In a physical model, the domain also depends on when the model applies.

Worked example · Voltage over time

A sensor model gives V(t) = 1 + 2t volts, where t is time in seconds and 0 ≤ t ≤ 3. Substituting t = 2 gives V(2) = 5 V. The domain is [0, 3] seconds. The range is [1, 7] volts. The graph rises by 2 volts every second.

V(t)7103t (s)

Read functions in four forms

A formula, table, graph, and verbal rule can describe the same relationship. For the voltage model, the table contains (0, 1), (1, 3), (2, 5), and (3, 7). The formula determines values between samples. A measurement table alone does not tell you how voltage behaves between readings. On a graph, a vertical line must meet the graph at most once for each input.

Domain restrictions and piecewise models

For real-valued formulas, exclude inputs that make a denominator zero and require a nonnegative expression under a square root. For example, f(x) = √(x − 2)/(x − 5) has domain [2, 5) ∪ (5, ∞). The root allows x = 2, but the denominator excludes x = 5.

A sensor clips a voltage that exceeds 5 V. On 0 ≤ t ≤ 3, its reading is 1 + 2t for t < 2 and 5 for t ≥ 2. Plot the rising segment up to (2, 5) and the flat segment from there to (3, 5). The second rule supplies the value at the boundary. The range is [1, 5] V, and a reading of 5 V no longer identifies a unique time.

Transform, compose and invert

In g(x) = a f(x − h) + k, move the graph horizontally by h, multiply its output by a, then shift vertically by k. A negative a reflects the graph vertically before the vertical shift. For f(x) = x², g(x) = −2(x − 3)² + 4 has vertex (3, 4), opens downward, and has range (−∞, 4].

A calibration stage converts voltage to a count using C(v) = 100v. With the original sensor model, (C ∘ V)(t) = C(V(t)) = 100 + 200t. Composition requires the inner output to be in the outer domain. To recover time from the unclipped voltage, solve v = 1 + 2t for t: the inverse is (v − 1)/2 for 1 ≤ v ≤ 7. An inverse function exists on a domain where distinct inputs give distinct outputs. The clipped sensor has no inverse on the full interval because different times can yield 5 V.

Common mistakes

  • Distinguish f(2), the output at 2, from 2f(x), twice the output at x.
  • Check both the formula restrictions and the physical measurement interval.
  • Test the input and output intervals when finding an inverse. The inverse function and reciprocal have different meanings.

Practice and check

A second model gives V(t) = 8 − 2t on 0 ≤ t ≤ 4. Find V(1), state the domain and range, and sketch the graph.

Show solution

V(1) = 6 V. The domain is [0, 4] seconds and the range is [0, 8] volts. The graph is a line segment from (0, 8) to (4, 0).

Challenge: let u(t) = t − 2 and g(x) = √x. Find the domain of g(u(t)) on the measurement interval [0, 5] and find its range.

Show challenge solution

The inner output t − 2 must be nonnegative, so intersect t ≥ 2 with [0, 5] to get [2, 5]. On this interval √(t − 2) increases from 0 to √3, so the range is [0, √3].

Quick check · 5 minutes

The same sensor records 4 V and 6 V at exactly t = 2 s. Can those two records describe a single function of time for the sensor? Explain.

Show answer

No. A single-valued function of time can assign only one voltage to t = 2. Check the measurements or include another input that distinguishes them.

OpenStax · Review of Functions

Virtual lab · Calibrate a voltage sensor

Predict a sensor reading and identify the clipping boundary.

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Procedure

The model is V(t) = min(1 + 2t, 5) V for 0 ≤ t ≤ 3 s. Predict the readings at 1, 2, and 3 s. Change the time below and record the outputs. Explain why a 5 V reading cannot identify a unique time.

Expected results and answer check

At 1, 2 and 3 seconds the outputs are 3, 5 and 5 V. The model clips from t = 2 s onward, so multiple inputs share the same output.

This model omits sensor noise and calibration error.

GeoGebra Graphing Calculator · Graph y = min(1 + 2x, 5) and locate the corner.

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